Mathematics I (Calculus) 2075

Given,

Now,

Now, for sketching graph  calculating y = f(x) for different values of x

Graph:

Given

Now,

Here,

Hence,  doesn’t exist.

Given

For domain, for all real values of x, f(x) exist. So, domain is set of all real number i.e. domain is  

For Graph, calculating the values of y = f(x) for different values of x;

Plotting these points on graph we get:

Given

And lines:

y =1, y = 2

The given curve is the parabola and the sketch of the given curve is

Required area 

We need to find derivatives of f(x) = cos x, so

Therefore, Maclaurin series for cos x  is

Since the cosine function and all the derivatives of cosine function have absolute value less than or equal to 1. So, by Taylor’s inequality

Now,

i.e.

 for all values of x.

This implies that the series converges to cosx for every value of x.

The problem of finding a function y of x when we know its derivative and its value y₀ at a particular point x₀ is called an initial value problem.

Given,

Comparing given equation with  we have

P =2 and Q=3

Now,

Applying the initial condition y(0)=1

Applying this value, we have:

The sphere of radius a can be obtained rotating the half circle graph (semi-circle) of the function

 about the x-axis. 

The volume V is obtained as follows:

by the symmetry about the y-axis,

Here

As  we get  form.

So, set  where m is some constant value then,

Along  we observe  and we get

And at m =1,

Thus,

So, the limit does not exist.

Given,

Now,

A function f is continuous from the right at a number a if  and f is continuous from the left at a if  .

A function f is continuous on an interval if it is continuous at every number in the interval. If f is defined only one side of an end point of the interval, we understand continuous at the end point to mean continuous from the right or continuous from the left.

Given,

Let  then

Which shows that f(x) is continuous at .

For the end points i.e. x=-1

Which shows that f(x) is continuous at the left end point x = -1

Similarly for the end point x=1

Which shows that f(x) is continuous at the right end point x = 1.

Hence f(x) is continuous at [-1,1]

Given,

f(x) = x³-3x+2 

Since, f(x) = x³-3x+2  is continuous on [-1, 2] and f’(x) = 3x²-3 so, differentiable on (-1, 2).

Thus f(x) = x³-3x+2 satisfy the both conditions for mean value theorem. So, there exist  such that

Clearly, 

Hence, mean value theorem satisfied.

Given,

By Newton’s method we have

When x₁=2

Then

Here

Take,

Put Then  So that

Thus, form (i)

Given,

The characteristics equation of given differential equation is

Here the roots are real and equal.

The general solution is

Given,

Here,

Now;

Therefore, by n^th term test for divergence, the given series is divergent.

Given

a = (4 , 0, 3)

b = (-2, 1, 5)

Now,

Given,

Now,

Differentiating w.r.to x

Again,

Differentiating w.r.to y

Given,

And, let

By method of Lagrange’s multiplier, for some scalar 

This implies,

This gives

From (ii) we have x = 0 or λ = 1. If x = 0, then (i) gives y = ±1. If λ = 1, then y = 0 from (iii), so then (i) gives x = ±1. Therefore, f has possible extreme values at the points (0, 1), (0, −1) (1, 0), and (−1, 0). Evaluating f at these four points, we find that

f(0, 1) = 2

f(0, −1) = 2

f(1, 0) = 1

f(−1, 0) = 1

Therefore, the maximum value of f on the circle x ² + y ² = 1 is f(0, ±1) = 2 and the minimum value is f(±1, 0) = 1.